2155 - When U are gone
When you walk away
I count the steps that you take
Do you see how much I need you right now?
——Avril《When you are gone》
有三位恋人（Alpha,Bob,Cindy），他们都在遥远的相望。（这………………）
他们都在说
离开你我度日如年，
我希望你坐飞机早些回来。
给出你我之间的路程，
和飞机每天能飞的路程。
快告诉我，你几天后能回来。
我要整数，不会有人说我在1.618346天后能看到你吧？！1.1天也是，只要大于1天，那么都算成两天！
题目输入
输入有且只有一组，这一组包含六个数，分成3行， 每组包括a,b两个数字(a!=0 && b!=0)。
题目输出
输入完三行后，输出三行信息，表示他们y天能够回来，注意单复数。
xx will be back in y day(s).
输入/输出样例
题目输入
4000 900 1800 900 2000 999
题目输出
Alpha will be back in 5 days. Bob will be back in 2 days. Cindy will be back in 3 days.
C语言解答
#include<stdio.h> int main(void) {int a,b,c,i; int k[10],j[10]; a=b=c=0; for(i=1;i<=3;i++) scanf("%d%d",&k[i],&j[i]); a=k[1]/j[1]; if(k[1]%j[i]!=0) a=a+1; b=k[2]/j[2]; if(k[2]%j[2]!=0) b=b+1; c=k[3]/j[3]; if(k[3]%j[3]!=0) c=c+1; if(a==1) printf("Alpha will be back in %d day.\n",a); else printf("Alpha will be back in %d days.\n",a); if(b==1) printf("Bob will be back in %d day.\n",b); else printf("Bob will be back in %d days.\n",b); if(c==1) printf("Cindy will be back in %d day.\n",c); else printf("Cindy will be back in %d days.\n",c); }
C++解答
#include<stdio.h> int main() { int a1,b1,a2,b2,a3,b3,c1,c2,c3; scanf("%d%d%d%d%d%d",&a1,&b1,&a2,&b2,&a3,&b3); if(a1%b1==0)c1=a1/b1; else c1=a1/b1+1; if(a2%b2==0)c2=a2/b2; else c2=a2/b2+1; if(a3%b3==0)c3=a3/b3; else c3=a3/b3+1; if(c1!=1) printf("Alpha will be back in %d days.\n",c1); else printf("Alpha will be back in %d day.\n",c1); if(c2!=1) printf("Bob will be back in %d days.\n",c2); else printf("Bob will be back in %d day.\n",c2); if(c3!=1) printf("Cindy will be back in %d days.\n",c3); else printf("Cindy will be back in %d day.\n",c3); return 0; }